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I need to calculate this definite integral: $$\int_\frac{\pi}{6}^\frac{5\pi}{6}\sqrt{(\sin(t)\cos(t))^2}\,dt$$ I can`t uncover this root, because of interval of x. Then what should I do? Help please.

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  • $\begingroup$ One may recall that $\sqrt{x^2}=|x|$. $\endgroup$ – Olivier Oloa Dec 15 '17 at 19:40
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    $\begingroup$ $$\frac{1}{2}\int_{\pi/6}^{5\pi/6}\left|\sin(2t)\right|\,dt=\frac{1}{4}\int_{\pi/3}^{5\pi/3}\left|\sin\theta\right|d\theta $$ is not that difficult to compute. $\endgroup$ – Jack D'Aurizio Dec 15 '17 at 19:41
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$$\int_\frac{\pi}{6}^\frac{5\pi}{6}\sqrt{(\sin(t)\cos(t))^2}\,dt= \int_\frac{\pi}{6}^\frac{5\pi}{6}|\sin(2t)|/2\,dt= {1\over 4}\int_\frac{\pi}{3}^\frac{5\pi}{3}|\sin(x)|\,dx= {1\over 4}\int_\frac{\pi}{3}^\frac{5\pi}{3}\sin(x)\,dx$$

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Hint:

Additivity of the integral w.r.t. intervals (a.k.a. Chasles' relation) is your friend.

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